The vapor pressure of pure water at 25 °C is 23.8 torr.What is the vapor pressure (torr)of water above a solution prepared by dissolving 18.0 g of glucose (a nonelectrolyte,MW = 180.0 g/mol)in 95.0 g of water?
A)24.3
B)23.4
C)0.451
D)0.443
E)23.8

Question

Quizplus · Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's Law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. First, calculate the moles of glucose and water. Moles of glucose = 18.0 g / 180.0 g/mol = 0.1 mol. Moles of water = 95.0 g / 18.0 g/mol = 5.28 mol. The mole fraction of water = 5.28 / (5.28 + 0.1) = 5.28 / 5.38 ≈ 0.981. The vapor pressure of water above the solution = 23.8 torr * 0.981 ≈ 23.4 torr.

The Vapor Pressure of Pure Water at 25 °C Is